Analysis of Covariance
Comparison of Head-Lengths for Trout from 2 Lakes
Source: C. McC. Mottley (1941). “The Covariance Method of Comparing the Head-Lengths of Trout From Different
Environments,” Copeia, Vol.1941, #3, pp.154-159.
Analysis of Covariance
• Goal: To Compare 2 or More Group/Treatment Means for One Variable, after controlling for one or
more concomitant variables/Covariates
• Procedure: Fit a Regression Model Relating Response to Group (Using Dummy Variables) and
Covariate(s)
• Adjusted Means: Fitted Values from regression for each group evaluated at common value(s) of
covariate(s), typically at the mean value(s)
• Treatments can be combinations of factors in a Factorial Structure
• Model:




Yij  i   1 X 1ij  X 1..     k X kij  X k ..  eij
Example – Fale Trout Heads from 2 Lakes (Environments)
•
•
•
•
•
•
•
Units: Female Fish Sampled from Lakes
Response Variable: Head Length (mm)
Groups: Environments (Kootenay Lake, Wilson Lake)
Covariate: Standard-Length (mm)
Samples: n1 = 50 from Kootenay, n2 = 10 from Wilson
Transformation: Base10 logarithms of Head-Length and Fish-Length Taken for Biological Reasons
Summary Statistics:
Lake
Head Mean
Head SD
Fish Mean
Fish SD
Correlation
Kootenay
1.7939
0.1761
2.4488
0.1731
0.9884
Wilson
1.7431
0.0750
2.3821
0.0806
0.9814
Overall
1.7854
0.1642
2.4377
0.1628
0.9875
T-test – Ignoring Covariate
Step 1 : Test for Equal Variance (  0.05) :
H 0 :  12   22
T .S . : Fobs
H A :  12   22
S12 (0.1761) 2
1
 2 

5
.
519

 0.181
S 2 (0.0750) 2
Fobs
Decision Rule : Conclude Variances are Equal (Do not reject H 0 ) unless :
Fobs  F.025, 49,9  3.475   12   22
1
 F.025,9, 49  2.387   12   22
Fobs
 Conclude variances differ, Do not Pool Variances and use degrees of freedom (Satterthw aite) :
2
df 



2
 S12 S 22 
 (0.1761) 2 (0.0750) 2 
  



n
n
50
10
2 
 1



(0.0750) 2 10
S12 n1
S 22 n2   (0.1761) 2 50
 


50  1
10  1
n1  1
n2  1  

 


 

(.001182) 2

 32.6  32
 4.29268E - 08


T-test – Ignoring Covariate
Step 2 : Test for Equal Means (  0.05) :
H 0 : 1   2
H A : 1   2
T .S . : tobs 
y1  y 2
S12 S 22

n1 n2

1.7939 - 1.7431
(0.1761) 2 (0.0750) 2

50
10
R.R. : tobs  t.025,32  2.04
P - value : 2 P t   1.47   .1513
 Do not conclude means differ.

 0.0508
 1.47
0.0344
Note : z.025  1.96
Note : 2 P z  1.47   2(.0708)  .1416
Adjusting for Body Length




Model : Yij  i   X ij  X ..  eij  1 K ij   2Wij   3 X ij  X ..  eij
i  1,2 j  1,..., ni
where :
Yij  Log 10 (Head  Length) for jth fish from Lake i
K ij  1 if Kootenay Lake (i  1), 0 otherwise
Wij  1 if Wilson Lake (i  2), 0 otherwise
X ij  Log 10 (Body  Length) for jth fish from Lake i
X ..  2.4377
Wilson
Kootenay
BodyLng
Coefficients
1.7988
1.7827
1.0021
Standard Error
0.00818
0.00363
0.02073
t Stat
219.9
491.3
48.3
P-value
0.0000
0.0000
0.0000
Adjusted Means
• Goal: Compare Means Head Lengths after adjusting for Body Lengths
• Start with the Unadjusted Mean, and Subtract off the product of the slope from the regression and the
difference between the group mean and the overall mean for the covariate. Effect of adjustment (assuming
positive slope):
– Adjust down for groups with high levels of covariate
– Adjust up for groups with low levels of covariate
Y
Adj
i
^

 Y i.   X i.  X ..

Trout Head-Length Data
Y 1.  1.7939
X 1.  2.4488
Y 2.  1.7431 X 2.  2.3821
^
  1.0021 X ..  2.4377
Y
Adj
1.
 1.7939  1.0021(2.4488  2.4377)  1.7828
Y
Adj
2.
 1.7431  1.0021(2.3821  2.4377)  1.7988
Plot of Regression and Means (Adjusted and Unadjusted)
2.2000
2.1000
Head Length
2.0000
Wilson
Kootenay
Wilson-Unadj
Kootenay-Unadj
Wilson-Adj
Kootenay-Adj
1.9000
1.8000
1.7000
1.6000
1.5000
2.25
2.3
2.35
2.4
2.45
2.5
Body Length
2.55
2.6
2.65
2.7
2.75
Testing for Lake Effect, Controlling for Body Length
•
•
•
•
Null Hypothesis: No difference in mean head lengths for 2 lakes, controlling for Body Length (12)
Alternative Hypothesis: Lake Means Differ (1≠2)
Step 1: Fit Full Model Under HA and obtain SSE(F), dfE(F)
Step 2: Place Restriction under H0 and fit model forcing 12 and obtain SSE(R) and dfE(R) :


Reduced Model : Yij  1Wij  1 K ij   3 X ij  X ..  eij




 Yij  1 Wij  K ij    3 X ij  X ..  eij  1U ij   3 X ij  X ..  eij
where : U ij  Wij  K ij
Then : The test Statistic is :
Fobs
 SSE ( R)  SSE ( F ) 
 df ( R)  df ( F ) 
E

 E
MSE ( F )
Trout Head Size Data
ANOVA for Full Model:
Regression
Residual
Total
ANOVA for Reduced Model:
df
3
57
60
SS
192.8131702
0.037372459
192.8505427
df
SS
MS
2
192.8111
96.40553
Residual
58
0.03949
0.000681
Total
60
192.8505
Regression
MS
64.27105674
0.000655657
 SSE ( R)  SSE ( F )   0.03949  0.03737 
 df ( R)  df ( F )  
 .00212
58  57
E
E



TS : Fobs 


 3.23
MSE ( F )
0.000656
.000656
RR : Fobs  F.05,1,57  4.010
P - Value : PF  3.23  .0776